Step 1: Analyze the contents of each box.

Let's list the number of balls of each color in the three boxes and the total number of balls in each box.

• Box B₁:

  • White (W): 1
  • Red (R): 3
  • Black (B): 2
  • Total balls in B₁: $1 + 3 + 2 = 6$

• Box B₂:

  • White (W): 2
  • Red (R): 3
  • Black (B): 4
  • Total balls in B₂: $2 + 3 + 4 = 9$

• Box B₃:

  • White (W): 3
  • Red (R): 4
  • Black (B): 5
  • Total balls in B₃: $3 + 4 + 5 = 12$

Step 2: Define the event.

We need to find the probability that all 3 drawn balls (one from each box) are of the same color. This can happen in three mutually exclusive ways:

1. All three balls are white (WWW).
2. All three balls are red (RRR).
3. All three balls are black (BBB).

The total probability will be the sum of the probabilities of these three events.

Let $P(E)$ be the required probability. $P(E) = P(\text{all are White}) + P(\text{all are Red}) + P(\text{all are Black})$

Step 3: Calculate the probability for each case.

The draws from the three boxes are independent events.

Case 1: All three balls are White.

The probability of drawing a white ball from B₁ is $P(W_1) = \frac{1}{6}$. The probability of drawing a white ball from B₂ is $P(W_2) = \frac{2}{9}$. The probability of drawing a white ball from B₃ is $P(W_3) = \frac{3}{12}$.

So, the probability that all three are white is: $P(WWW) = P(W_1) \times P(W_2) \times P(W_3) = \frac{1}{6} \times \frac{2}{9} \times \frac{3}{12} = \frac{6}{648}$

Case 2: All three balls are Red.

The probability of drawing a red ball from B₁ is $P(R_1) = \frac{3}{6}$. The probability of drawing a red ball from B₂ is $P(R_2) = \frac{3}{9}$. The probability of drawing a red ball from B₃ is $P(R_3) = \frac{4}{12}$.

So, the probability that all three are red is: $P(RRR) = P(R_1) \times P(R_2) \times P(R_3) = \frac{3}{6} \times \frac{3}{9} \times \frac{4}{12} = \frac{36}{648}$

Case 3: All three balls are Black.

The probability of drawing a black ball from B₁ is $P(B_1) = \frac{2}{6}$. The probability of drawing a black ball from B₂ is $P(B_2) = \frac{4}{9}$. The probability of drawing a black ball from B₃ is $P(B_3) = \frac{5}{12}$.

So, the probability that all three are black is: $P(BBB) = P(B_1) \times P(B_2) \times P(B_3) = \frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} = \frac{40}{648}$

Step 4: Calculate the total probability.

The total probability is the sum of the probabilities of the three mutually exclusive cases. $P(E) = P(WWW) + P(RRR) + P(BBB)$ $P(E) = \frac{6}{648} + \frac{36}{648} + \frac{40}{648}$ $P(E) = \frac{6 + 36 + 40}{648} = \frac{82}{648}$

Step 5: Compare with the options.

The calculated probability is $\frac{82}{648}$, which matches option A.